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Calculate the amount of benzoic acid `(C_(6)H_(5)COOH)` required for preparing `250mL` of `0.15 M` solution in methanol.

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`"Molarity (M)"=("Mass of solute/molar mass")/("Volume of solution i litres")`
M=0.15 M =0.15 mol `L^(-1)`,
Molar mass of solute= `7xx12+6xx1xx2xx16=122 h mol^(-1)`
Volume of solution = 250 mL = 0.25 L
`(0.15 mol L^(-1))=("Mass of solute")/((122 g mol^(-1))xx(0.25L))`
Mass of solute= `(0.15 g mol L^(-1)xx(0.25 L)=4.575 g`

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